Question

Let S be the infinite sum given by $S=\sum _{n=0}^{\infty }\frac{a_n}{{10}^{2n}}$, where $(a_n{)}_{n\ge 0}$ is a sequence defined by $a_0=a_1=1$ and $a_j=20a_{j-1}$ for $j\ge 2$. If $S$ is expressed in the form $\frac{a}{b}$, where $a,b$ are coprime positive integers, than $a$ equals.

• A. $60$
• B. $75$
• C. $80$
• D. $81$

Answer 1

The correct option is D $81$

Given, $S={\sum }_{n=0}^{\infty }\frac{a_n}{{10}^{2n}}$

Expanding the summation, we get

$S=\frac{a_0}{1}+\frac{a_1}{{10}^2}+\frac{a_2}{{10}^4}+\frac{a_3}{{10}^6}+\frac{a_4}{{10}^8}+\dots$

On substituting the given values, we have

$S=1+\frac{1}{{10}^2}+\frac{20}{{10}^4}+\frac{{20}^2}{{10}^6}+\frac{{20}^3}{{10}^8}+\dots$

$S=1+\frac{1}{{10}^2}+\frac{2}{{10}^3}+\frac{2^2}{{10}^4}+\frac{2^3}{{10}^5}+\dots$

$S=1+\frac{1}{{10}^2}\{1+\frac{2}{10}+\frac{2^2}{{10}^2}+\frac{2^3}{{10}^3}+\dots \}$

$S=1+\frac{1}{{10}^2}\{1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots \}$

Since the terms in the bracket form an infinite geometric progression, we can sum them as

$S=1+\frac{1}{{10}^2}\left\{\frac{1}{1-\frac{1}{5}}\right\}$

$S=1+\frac{1}{{10}^2}\left\{\frac{5}{4}\right\}$

$S=1+\frac{5}{400}$

$S=\frac{405}{400}$

As we have coprime numerator and denominator requirement,

$\frac{a}{b}=\frac{81}{80}$

Hence, the required answer is $81$.